Question

A heavy object is dropped from a vertical distance of 12.0 m above the ground .What is the speed of the object as it hits the ground . Take g = 9.8 m/s^2

Answer 1

Answer:

15.33 m/s

Explanation:

Given :

• Height of the heavy object from the ground = 12.0 metres = s

• Gravitational force = 9.8 m/s² = a

• Initial velocity = u = 0 m/s (As the heavy object starts from rest when dropped)

To find :

• The final velocity of the heavy object = V

Using the third equation of motion :

• V²-u²=2as

V²-0²=2×9.8×12

V²-0=19.6×12

V²=235.2

V=15.33 m/s

The speed with which the object hits the ground is equal to 15.33 m/s

Answer 2

AnswEr :

Given that height from which object is dropped is 12.0 m. Gravitational acceleration (g) = 9.8 m/s²

We have to find speed of object as it hits the ground.

We have,

➡ Height (h) = 12 m

➡ Acceleration (g) = 9.8 m/s²

➡ Initial speed (u) = 0 [As it drops freely]

Now we will use 3rdd equation of motion :

→ v² - u² = 2gh

Substituting values,

→ v² - 0² = 2 × 9.8 × 12

→ v² - 0 = 235.2

→ v² = 235.2

→ v = √235.2

→ v = 15.34 m/s

So, speed of object as it hits ground = 15.34 m/s [Answer] ✔